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User:Yunzhong Hou - Wikipedia, the free encyclopedia

User:Yunzhong Hou

From Wikipedia, the free encyclopedia

n3

\frac{1}{\tan c}

\sqrt{2 + 4/5 \frac{3}{2}}

\sum_{k=1}^N k^2 + k^3 + k^4 + ...

1 (2) + 2 (3) + 3 (4) + ... + n (n+1) = \frac{n ( n + 1 )}{2}

1(2) + 2(3) + 3(4) + ... + n(n + 1) = ?

an = n(n + 1)

Sn

S_n = \frac{n(n+1)(n+2)}{3}

1(2) = \frac{1(1+1)(1+2)}{3} = 2

1(2) + 2(3) + ... + n(n+1) = \frac{n(n+1)(n+2)}{3}

1(2) + 2(3) + ... + n(n+1) + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3}

1(2) + 2(3) + ... + n(n+1) + (n+1)(n+2) = \frac{n(n+1)(n+2)}{3} + (n+1)(n+2)

= \frac{n(n+1)(n+2)+3(n+1)(n+2)}{3}

= \frac{(n+1)(n+2)(n+3)}{3}

12 + 22 + ... + n2 = (1(2) + 2(3) + ... + n(n + 1)) + (1 + 2 + ... + n)

= \frac{(n-1)(n)(n+1)}{3} + \frac{n(n+1)}{2}

= \frac{1}{3} n^3 - \frac{1}{2} n^2 + \frac{1}{6} n

S_n = \frac{n(n+1)(n+2)(n+3)}{4}

1(2)(3) = \frac{1(1+1)(1+2)(1+3)}{4} = 6

1(2)(3)+2(3)(4)+...+n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4}

= \frac{n(n+1)(n+2)(n+3)}{4}+(n+1)(n+2)(n+3)

= \frac{n(n+1)(n+2)(n+3)+4(n+1)(n+2)(n+3)}{4}

= \frac{n(n+1)(n+2)(n+3)(n+4)}{4}

= \frac{(n-1)(n)(n+1)(n+2)}{4}+ \frac{n(n+1)}{2}

= \frac{1}{4}n^4- \frac{1}{2}n^3+ \frac{1}{4}n^2

= \frac{1}{4}n^4- \frac{1}{2}n^3+ \frac{1}{4}n^2

S_n = \frac{n(n+1)(n+2)(n+3)(n+4)}{5}

1(2)(3)(4) = \frac{1(1+1)(1+2)(1+3)(1+4)}{5} = 24

1(2)(3)(4)+2(3)(4)(5)+...+n(n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)(n+4)}{5}

1(2)(3)(4) + 2(3)(4)(5) + ... + n(n + 1)(n + 2)(n + 3) + (n + 1)(n + 2)(n + 3)(n + 4)

= \frac{n(n+1)(n+2)(n+3)(n+4)}{5}+(n+1)(n+2)(n+3)(n+4)

= \frac{n(n+1)(n+2)(n+3)(n+4)+5(n+1)(n+2)(n+3)(n+4)}{5}

= \frac{n(n+1)(n+2)(n+3)(n+4)(n+5)}{5}

\frac{1}{5}n^5- \frac{1}{2}n^4- \frac{1}{3}n^3- \frac{1}{30}n



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